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8/6/2019 Stability Part 1
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Understand the importance of system
stability
Identify stable & unstable systems
State the characteristic function,
characteristic equation, system zeroes and
poles of a system
Analyse stability in the s-plane
Objectives
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On 7 Nov 1940, at about 10 a.m. in
TACOMA, Washington, USA something
happened .
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Think about the Tacoma story.
Find out more about the story and answer the following:
What happened in Tacoma on 7 Nov 1940? What could be the cause of that?
How is it related to what you are learning?
What have you learned ?
Your Tasks
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Concepts on System Stability
What is stability of a system?
Why is it important?
What are some examples?
Can we know if a system is stable from its
mathematical model? If so, how?
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A control system is stable if the output eventually settles
down to a finite value due to a set-point change, load change
or disturbance.
SystemInput
r(t)
Output
c(t)
SystemStable-value)finite(a)(lim!
gp
tct
SystemUnstable-value)(infinite)(lim g!gp
tct
What is Stability ?
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The following are time-response of outputs of stable systems
C(t)
t0
C(t)
t0
SystemStable-value)finite(a)(lim !gp
tct
a a
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The following are time-response of outputs of unstable systems
C(t)
t0
C(t)
t0
Systemnstable-value)(in inite)(lim g!gp
tct
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A special case is the marginally stable system where the
output oscillates continuously.
C(t)
t0
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Robotic Arm
Nuclear Power Plant
Missile
Oscillator
Tacoma Narrows Bridge **
Why is stability important?
What are some examples?
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There are many ways of determining the stability of control
systems :
a. Based on Time Response
b. Based on Roots of Characteristic Equation
c. Routh- Hurwitz Stability Criterion
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Stability based on Time Response
The procedures to determine the stability of a closed-loop
control system are:
a. Find the closed-loop transfer function
b. Find the output c(t)
c. Find c(t) as t E
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Example 1
C(s)R(s)+
- 11
s
inputstepafors
1R(s)*
2
11
2
1
)()(
2
1
)(
)(
!
v!
!
!
ss
ssRsC
ssR
sC
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1
221
)(
)(
kc
ekktct
!g
!
Since the output stabilises at k1, the system is stable
Example 1...
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Example 2
C(s)R(s) +
- )5.0)(2.0(1
ss
)5.0)(2.0(1
11
)5.0)(2.0(1
1)()(
)5.0)(2.0(1
1
)(
)(
v!
!
!
sss
ssss
sss
s
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)sin()()( 135.0
211 J!! tekksCLtc t
system is stable.1)(lim ktct
!
gp
Example 2...
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Example 3
C(s)R(s) +
- )2.0)(5.0(1
ss
)2.0)(5.0(1
11
)2.0)(5.0(1
1)()(
)2.0)(5.0(1
1
)(
)(
v!
!
!
sss
ssss
sss
s
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)94.0sin()()( 115.0
211 J!! tekksCLtc t
gp
gp )(limtc
t
system is unstable.
Example 3...
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Example 4
C(s)+R(s)
- 31
2s
4
11
4
1
)()(
4
1
)(
)(
2
2
2
v!
!
!
ss
ssRsC
ssR
sC
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Sustained Oscillation
)2sin()()( 1211 J!! tkksCLtc
p
gp
)(lim tct
Example 4...
System is marginally stable
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Stability based on Roots of Characteristic Equation
The Characteristic Equation is formed from the denominator
of the closed-loop transfer function.
Consider the closed-loop control system :
C(s)R(s)+
-G(s)
H(s)
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0)()(1
:issystemtheo( )quationsticharacteriThe
)()(1
)(
)(
)(
:issystemor thisunctiontrans erloopclosedThe
!
!
sHsG
sHsG
sG
s
s
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Determination of the Stability
The following are steps to determine the stability of a
closed-loop control system using the Characteristic Equation:
1. Find the Characteristic Equation of the system.
2. Determine the roots of the Characteristic Equation.
3. Locate the roots on the S-plane.
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Determination of the Stability (continued..)
If the characteristic equation of the system has a
pair of complex-conjugate roots, the system response
will be oscillatory.
System is marginally stable if there is
at least one pair ofcomplex-conjugate
roots on the j[ axis.
4. System is stable ifall the root are in the
Left-Hand-Side of the S-plane.
System is unstable ifone or more rootsare in the Right-Hand-Side of the S-plane.
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We will now use the CE to determine the stability of
Example 1 to Example 4 :
The CE for Example 1 is :
position.2-at thecrossabydrepresenteisThis
-2sisrootthei.e.
02
!
!s
x
-2
Since the root lies on the LHS of the s-plane the
system is stable.
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In Example 2
)5.0)(2.0(1
1
)(
)(
!
sss
s
Roots :s1 = - 0.35 + j, s2 = - 0.35 - j
j0.661.25-
0)5.0)(2(1:
s!
! ssCE
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-0.3
x
x
j
-j
Since the roots are on the LHS of the s-plane, the
system is stable.
However, as the roots are complex, the transient period
is oscillatory.
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1)5.0)(2.0(
1
)(
)(
!
sssR
sC
In Example 3
CE : (s + 0.2)(s - 0.5) + 1 = 0
Roots of CE are located on RHS System is unstable
x
x
0.94j
-0.94j
0.15
Roots : s1 = 0.15 + 0.94j, s2 =0.15 - 0.94j
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4
1
)(
)(2
!
ssR
sC
In Example 4
CE : s2 + 4 = 0
x
x
2j
-2j
Roots : s1 = 2j, s2 = - 2jRoots of CE are on the j[ axis System is marginally stable
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Click anywhere on this slide to for an interesting
program that shows the relationship between the
pole positions(roots of the Characteristic equation
and the transient response.
Have fun.